Is it more efficient to drive fast uphill? I know that for a rocket escaping the atmosphere, it's not efficient to travel slowly because even staying stationary consumes a lot of fuel. Does the same apply to a vehicle traveling uphill? In other words, is it more efficient for a car/bike/runner to accelerate and go over the hill quickly because energy is exerted during the climb even to "keep the vehicle stationary" and not just to travel the height and distance?
 A: update: this answer is all wrong. See comments.
Ignoring optimal engine RPM and accelerations, assume there is no rolling resistance so that the only frictional force is air resistance.
$$F_{air} = cv^2$$
where $c$ is the drag coefficient for the given car and air conditions. The energy lost to it is:
$$E_{air} = dF_{air} = dcv^2$$ where $d$ is distance travelled. The net work energy to reach the top of a hill is $E_{height} = mgh$ where $h=d\sin(A)$.
The energy the engine must provide is:
$$E_{air} + E_{height} = dcv^2 + mgd  \sin(A)$$
Assume the most efficient $v$ is a constant so we can simply say $d = t/v$ where $t$ is time that $d$ was travelled. To find the $v$ that minimizes $E_{air} + E_{height}$ we take the derivative with respect to $v$, set it to zero and solve:
$$0 = tc - mgt \sin (A)/v^2\implies v = \sqrt{\frac{mg \sin(A)}{c}}$$
We can measure $c$ by rolling the car downhill. Let V be the terminal speed which is when air resistance equals the force from gravity.
$$cV^2 = mg \sin(A)$$
Solving for $c$ and plugging into equation for $v$ gives
$$v =  V$$
A: So this is by no means easy to calculate and depends on MANY factors.
So first things is how efficiently can your engine turn energy into a give amount of force and how much force. For an engine this comes down to rotational speed and the Torque it can handle, for combustion engines that's much more complex than for general electric (DC) motors.

Okay that makes it quite complicated to deal with you'll see why in a bit. For now we'll just ignore engine efficiency and assume it's the same over the whole range, keep in mind I showed you an electric motor and not the more complex variant, a combustion engine.
Now when looking at this problem there's a total of 4 main forces, Roll resistance $F_r$, Air drag resistance $F_d$, Gravitational pull $F_g$ and the force needed to accelerate $F_a$. You're engine has to produce a $Torque = \frac{F_{sum}}{R_{wheels}}$ assuming you have no gearbox (true around gear 5 or so).
The easiest situation, let's assume that the roll resistance is more or less constant and small, this is probably true and let's also say that we look at lower speeds about 40 km/hr or lower and you have a sports car, so the coefficient of drag, $C_d$ is also low. In that case we can approximate $F_{sum} \approx F_g + F_a$.
Now there are two different time-values you need to calculate the energy needed. Let's call them $t_1$ and $t_2$. Where $t_1 = \frac{v_f}{acceleration}$ and $t_2 = \frac{distance_{adjusted}}{v_f}$, in which the adjusted distance, is the total distance minus the distance travelled during $t_1$, $d_{trav}=acceleration \cdot t_1^2$. Should you assume that $t_1 << t_2$ then you can assume that all that matters is the $t_2$ and with it $v_f$. Now this shows that even the distance you have to travel is important, but if you assume that the distance you have to travel is very large then the time to accelerate doesn't matter (under these conditions). Now it's very simple to calculate the energy needed which is simply the Power times the time. In this case $Torque=\frac{F_g}{R}$ and $w(angular.speed)=\frac{v_f}{R}$ and with $P(ower)=T(orque) \cdot w(angular.speed)$ we can then say the following:
$E(nergy)= P \cdot t_2= \frac{T \cdot w \cdot dist \cdot R}{w}= dist*F$
What does this mean, well good question, it shows that going faster doesn't improve things by 1 bit nor does going slower. To clarify things a bit further and where I think the misconception is in, Indeed the force needed to keep going at a specific speed "doesn't change" (in our simplified situation). That is definitely true, so if the force is the same we don't need to produce more force so we would expect less fuel consumption and this is the fallacy, the power needed is $P= T \cdot w$, so even though we spent less time applying the force, we will need more power specifically because a higher speed (thus $w$) means more power needed, even though the force is the same. So a higher speed is a shorter time and the same force, it does mean more power.
Now of course it's much more complicated when you start taking Air drag into account as the force increase with the square of your speed. With $F_d= C_1 \cdot v_f^2$ and here I take all the other factors as a constant called $C_1$ since in you're problem they will approximately be constant.
So if we again assume that accelerating takes very little time than $F_{sum} \approx F_d + F_g = C_1 \cdot v_f^2 + C_2$ and this is where it starts getting complicated as now we need to know the exact relation between the Torque, rotational speed, and Power. Generally speaking (perfect engine and all) $P=T \cdot w$. Here it becomes kind of messy but going faster is less efficiently.
Unfortunately in practice even this is not true, because you would have to take into account the power curve and efficiency curve of the engine, have fun I would say. So if you really want to solve this and get a better understanding, get the Torque vs angular speed graph of your engine, get the efficiency graph of your engine, use that to calculate and plot the power out and power in (power in is power out divided by efficiency). Finally for a specific car and angle you can also calculate the Torque vs angular speed, plot this line in the same graph. This let's you see which range of angular speeds is viable, then you look in that range where the minimum power in is and that's your most efficient speed.
As for a rocket well, that's a different story, especially since the more time you spent not doing anything useful, is fuel you're wasting or carrying along. Rockets have the notorious issue, where carrying a little bit of extra mass means carrying a lot more fuel.
Rockety stuff, The rocket fuel equation, What if? launch those idiots.
A: It depends on where the motor is most efficient. Typically you look at power output per fuel flow rate (Brake Specific Fuel Consumption) and for an internal combustion engine, the most efficient operation is almost full throttle at mid-low rpm.
Wikipedia has an example

Source: Motortechnische Zeitschrift MTZ 1/2005; Der neue Dreizylinder-Dieselmotor von Mercedes Benz für Smart und Mitsubishi
So to drive most efficiently on a long uphill road you need to be at the highest gear that puts the car near the 2100 rpm efficiency island, with about 90% throttle (relative distance of x-axis to the red dashed line).
If you get to the $206 \mbox{ (grams per second per horsepower)}$ island you will be operating at the most efficient way for this motor.
Each motor has their own BSFC map, and even electric, jet and other propulsion units will have a similar efficiency map. 
A: Keeping a rocket from falling requires continuous spending of energy in a form of burning fuel, while, under typical conditions, it takes no energy to keep an object on a slope of a hill.
So the transition from rockets hanging up in the air to cars and bicycles going up the hill is not quite justifiable. 
We can reasonably assume that, for each type of a vehicle and for each individual, there should be an optimal speed or range of speeds for going up a given hill at which the efficiency would be maximized, but that speed would depend more on the specifics of engines and muscles and on the air resistance than on the energy required to "keep the vehicle stationary" on the slope.
A: Generally, yes the faster you go the less energy you spend fighting gravity, but the more you spend overcoming resistance. At the vertical extreme, the faster the better. But at the horizontal extreme, the closer to the most economical cruise speed the better. For any given vehicle up any given slope, there will be an optimum speed in between these two extremes.
But the most economical way to drive up a hill is to put your foot down a bit some way before the hill and hold it there while you accelerate to the bottom of the hill, then decelerate as you climb until you reach the top, then accelerate back up to cruising speed. By holding you foot steady, you avoid the nonlinear excesses of power which waste energy. A refinement is to begin putting your foot down sooner and more gently so you accelerate more economically, and to begin raising it likewise at the top.
A: Some preliminaries: there is no time limit, no specific initial conditions and there is no wind. The road is straight and empty. There is no speed limit and safety is of no concern. In short, this has to be cheap.
Make sure you arrive at the speed $v > \sqrt{2gh}$. You need an excess to compensate for rolling resistance and air resistance. At the foot of the hill, go out of gear and switch off the engine. If you did this with enough fingerspitzengefühl, you car will reach the top at speed very near zero. Wait until it rolls down the other side. Switch the engine back on, raise the rpm's to the same value and go into the same gear as at the start of the climb. The whole trip has cost you nothing more than some dissipation by air and rolling resistance.
